Là sao v bn?

chắc là test TA

A = -x^2 - 6x + 1

= -(x^2 + 6x + 9) + 10

= -(x + 3)^2 + 10 ≤ 10

GTLN của A là 10 khi x = -3.

Ta có: \(A=-x^2-6x+1\)

\(=-\left(x^2+6x-1\right)\)

\(=-\left(x^2+6x+9-10\right)\)

\(=-\left(x+3\right)^2+10\le10\forall x\)

Dấu '=' xảy ra khi x+3=0

=>x=-3

x=2

Ta có: \(x^3-3x^2+3x-2=0\)

=>\(x^3-3x^2+3x-1-1=0\)

=>\(\left(x-1\right)^3=1\)

=>x-1=1

=>x=2

Ta có: \(8x^3+36x^2+54x+27=0\)

=>\(\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3=0\)

=>\(\left(2x+3\right)^3=0\)

=>2x+3=0

=>2x=-3

=>\(x=-\frac32\)

8x^3 + 36x^2 + 54x + 27 = 0

→ (2x)^3 + 3(2x)^2 * 3 + 3(2x) * 3^2 + 3^3 = 0

→ (2x + 3)^3 = 0

→ 2x + 3 = 0

→ x = -3/2

Vậy x = -3/2.

Ta có: \(x^3-3x^2+3x-1=\left(3x+5\right)^3\)

=>\(\left(x-1\right)^3=\left(3x+5\right)^3\)

=>3x+5=x-1

=>3x-x=-1-5

=>2x=-6

=>x=-3

x^3 - 3x^2 + 3x - 1 = (3x + 5)^3

→ (x - 1)^3 = (3x + 5)^3

→ x - 1 = 3x + 5

→ -2x = 6

→ x = -3

Vậy x = -3.

Ta có: \(x^3-6x^2+12x-8=\left(2x+1\right)^3\)

=>\(\left(x-2\right)^3=\left(2x+1\right)^3\)

=>2x+1=x-2

=>2x-x=-2-1

=>x=-3

x^3 - 6x^2 + 12x - 8 = (2x + 1)^3

→ (x - 2)^3 = (2x + 1)^3

→ x - 2 = 2x + 1

→ -x = 3

→ x = -3

Vậy x = -3.